As in the book. Each of the other thermopotentials yield similar and useful relations. Let s just do the Legendre Transformations and list them all:

Transcription

1 hysics 430 Solutions to roblem Set 9 1. Schroeder roblem 5.1. Mixed derivatives are fun and useful. One of the most important things that we have yet to see is that thermodynamics INSISS that certain things, things that you might never expect to be related (say like the way temperature changes with volume at fixed entropy compared with the way pressure changes with entropy at fixed volume), are closely related, maybe even equal! his ability to connect seemingly disparate things is one of the most important parts of using thermodynamics. he Maxwell Relations here are derived from the four thermodynamic potentials, U(S,, N), F(,, N), G(,, N), and H(S,, N). In each case, there is an equation that we have for the total differential, which allows us to relate first partial derivatives for each function to functions of state. he most fundamental is the combined first and second law relation for the internal energy: du ds d dn From which we recognize that: U U ; S, N S, N he smoothness assumption about these functions then insists that the cross derivatives are equal. herefore, for the case of the internal energy, U U S S, N S, N S, N, N Or S S, N, N As in the book. Each of the other thermopotentials yield similar and useful relations. Let s just do the Legendre ransformations and list them all: F U S; df Sd d dn G F ; dg Sd d dn H U ; dh ds d dn So

2 hysics 430 roblem Set 9 Fall 009 F F S ;, N, N G G S ;, N, N H H ; S, N S, N ake the cross derivatives and you get: S, N, N S, N, N S How nice!, N S, N. roblem. As a follow-on to the previous problem, let s try an example where the cross derivative idea gets a bit of exercise:. Suppose that you are interested in the thermodynamic behavior of solids. You know that you re mostly interested in situations where the number of particles inside the solid is fixed, so you expect that you should be able to find internal energy and entropy functions of the form: U, and S, In the spirit of the early part of the course, you expect that you might be able to calculate these functions along specific paths if you had an equation of state and perhaps some heat capacities. So go out and measure them! You re going to find that for most solids a good equation of state is: 1 0 where, the coefficients and, are the thermal expansion coefficient and the isothermal compressibility. For this problem, assume that they are independent of temperature. Similarly, you ll find that the heat capacity at constant volume is:

3 hysics 430 roblem Set 9 Fall 009 C 3 Nk B Maybe these results are OK, but how do you really know? Fortunately, we have a way to cross check that things are at least internally consistent. Let s see how: a) Write down the total differential for the internal energy, assuming that it is indeed a function of temperature and volume. Do it first as a mathematician would, in terms of the partial derivatives of the respective functions with various things held constant. All I was looking for here was: U U du d d he first partial derivative is our old friend, the heat capacity at constant volume. At least one of the partial derivatives is familiar to you (which one?) and so you know exactly what it is (what is it?). However, what about this one?: U It looks a lot like the pressure, but it s not. Which derivative of the internal energy IS the pressure? Ah well, so anyway, this derivative isn t the pressure because you re holding the wrong thing constant, but we can relate it to things that you CAN relate to pressure. First, draw some inspiration from Eq. 1.45, where Schroeder wiggles out the heat capacity at fixed pressure by using the 1 st law to write the heat in terms of energy and work. ry to do the same thing here, writing the change of internal energy as heat plus work and write heat in terms of entropy and work in terms of pressure and change of volume. Show that: U Your are likely to need at least one of the Maxwell relations from the previous problem to make it to this final result. OK, so the inspiration I hoped you d get was to play wild and loose with the partial derivative by writing the change in internal energy as the heat plus the work: U S d S

4 hysics 430 roblem Set 9 Fall 009 Use a Maxwell Relation for the Helmholtz Free energy, and you have the final result: U b) Assuming that the internal energy is a smooth and well-behaved function, we expect that the order of differentiation should not matter. herefore, U U Use this fact to decide whether our data for the equation of state and the heat capacity are internally consistent. Are they consistent? he cross derivatives must be equal, so the heat capacity and the equation of state are forced to be related so that: U U C Both derivatives are zero, so at least in this way, the equation of state and the heat capacity are internally consistent. 3. Schroeder roblem In the early stages of the course, we discussed how you can use a few experimental results, heat capacities and the equation of state for example, to allow us to integrate along specific paths to find the internal energy function and the entropy function. his problem shows you that the heat capacities at constant volume and at constant pressure must be related in a very specific way. hey are NO independent. a) By now, you re all getting pretty good at seeing through these things. If entropy is a function of the temperature and volume, then: S S ds d d he first partial derivative, when multiplied by temperature will tell us the heat input for the small temperature change, so: C Q S

5 hysics 430 roblem Set 9 Fall 009 b) Really, we are playing with different ways to write the functional dependences. In the last part, we assumed that entropy is a function of temperature and volume. Still, perhaps it s easier to think of the volume as a function of the pressure and temperature, so that: d d d retty fun, right? OK, so if that s what the volume change is, then I suppose that we can write the differential entropy as: S S ds d d d herefore, the total differential is: S S S ds d d In this way of writing things, we have now a total differential of entropy, written as a function of temperature and RESSURE. herefore, thinking like a mathematician again, we must have that: S S S and S S he first relation, multiplied by temperature says that: S C C c) Use the Maxwell Relations to find: C C, N Use the results of roblem 1.46 (c) and the definition of isothermal compressibility from 1.46 (b) we have very quickly the desired result:

6 hysics 430 roblem Set 9 Fall 009 C C, N d) he volume change with temperature and the volume change with pressure for the ideal gas can be directly calculated from the ideal gas equation of state: 1 NkB 1 NkB SO NkB NkB Nk B his is the standard result for the difference in heat capacities for the ideal gas. e) You can see that equilibrium systems, where the thermal expansion and compressibility are both positive, the heat capacity at constant pressure must be larger than the heat capacity at constant volume. f) Here you get to put some actual numbers into the results above. We find For Water at 5C, =.57x10-4 K -1 and =4.5x10-10 a K Without going any further, you can see that until you choose a volume, it s hard to say what the difference in heat capacities might be. It s the usual problem with extensive quantities: he quantities depend on system size. Just because I know that water has a heat capacity of 1 cal/(gram C) or 4. Joules/(gram K) and that water has a mass of 1 gram/cm3, then I know that the heat capacity of water is around 4. Joules/K for a 1 cubic centimeter volume. herefore, let s see what the 1 cubic centimeter value would be. hat s 10-6 cubic meters, so we get:

7 hysics 430 roblem Set 9 Fall 009 C C J / gram. K hat s a fractional difference or around 1% of the total heat capacity for water. You could also do the comparison for a mole of water, or 18 grams. hen the heat capacity is around 80 J/K and it s still a 1% effect. By comparison, for mercury we have (again, for a cubic centimeter) 4 300K C C J / K cc of mercury is 13.6 grams, compared to 00 grams for a mole, so you d need 14.7 cc/mole. herefore, the molar difference is around 3.5 J/K compared with the molar heat capacity of mercury of around 5 J/K. You can see that for mercury the difference is more like 14%. It s larger for mercury. Much of this difference is due to the differences in compressibility: Water is nearly incompressible and it shows up in that you don t expand the environment much during heat addition, SO not much additional work and not much difference in heat capacities. g) retty clear that if is zero, that there is not heat capacity difference. However, finite values (and a finite compressibility too) will cause a difference in the heat capacities.